【刷题】SPOJ 705 SUBST1 - New Distinct Substrings

Given a string, we need to find the total number of its distinct substrings.

Input

T- number of test cases. T<=20; Each test case consists of one string, whose length is <= 50000

Output

For each test case output one number saying the number of distinct substrings.

Example

Input:

2

CCCCC

ABABA

Output:

5

9

Solution

后缀排序后

一个后缀 \(SA[i]\) 有 \(n-SA[i]+1\) 个子串，但后缀 \(SA[i]\) 与 \(SA[i-1]\) 有 \(height[i]\) 个字符相同，那么就有 \(height[i]\) 个子串一样，减去就是了

#include<bits/stdc++.h>
#define ui unsigned int
#define ll long long
#define db double
#define ld long double
#define ull unsigned long long
const int MAXN=50000+10;
char s[MAXN];
int T,n,m,SA[MAXN],rk[MAXN],cnt[MAXN],nxt[MAXN],height[MAXN];
template<typename T> inline void read(T &x)
{
	T data=0,w=1;
	char ch=0;
	while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
	if(ch=='-')w=-1,ch=getchar();
	while(ch>='0'&&ch<='9')data=((T)data<<3)+((T)data<<1)+(ch^'0'),ch=getchar();
	x=data*w;
}
template<typename T> inline void write(T x,char ch='\0')
{
	if(x<0)putchar('-'),x=-x;
	if(x>9)write(x/10);
	putchar(x%10+'0');
	if(ch!='\0')putchar(ch);
}
template<typename T> inline void chkmin(T &x,T y){x=(y<x?y:x);}
template<typename T> inline void chkmax(T &x,T y){x=(y>x?y:x);}
template<typename T> inline T min(T x,T y){return x<y?x:y;}
template<typename T> inline T max(T x,T y){return x>y?x:y;}
inline void GetSA()
{
	n=strlen(s+1),m=300;
	for(register int i=1;i<=n;++i)rk[i]=s[i];
	for(register int i=1;i<=m;++i)cnt[i]=0;
	for(register int i=1;i<=n;++i)cnt[rk[i]]++;
	for(register int i=1;i<=m;++i)cnt[i]+=cnt[i-1];
	for(register int i=n;i>=1;--i)SA[cnt[rk[i]]--]=i;
	for(register int k=1,ps;k<=n;k<<=1)
	{
		ps=0;
		for(register int i=n-k+1;i<=n;++i)nxt[++ps]=i;
		for(register int i=1;i<=n;++i)
			if(SA[i]>k)nxt[++ps]=SA[i]-k;
		for(register int i=1;i<=m;++i)cnt[i]=0;
		for(register int i=1;i<=n;++i)cnt[rk[i]]++;
		for(register int i=1;i<=m;++i)cnt[i]+=cnt[i-1];
		for(register int i=n;i>=1;--i)SA[cnt[rk[nxt[i]]]--]=nxt[i];
		std::swap(nxt,rk);
		rk[SA[1]]=1;ps=1;
		for(register int i=2;i<=n;rk[SA[i]]=ps,++i)
			if(nxt[SA[i]]!=nxt[SA[i-1]]||nxt[SA[i]+k]!=nxt[SA[i-1]+k])ps++;
		if(ps>=n)break;
		m=ps;
	}
	for(register int i=1,j,k=0;i<=n;height[rk[i++]]=k)
		for(k=k?k-1:k,j=SA[rk[i]-1];s[i+k]==s[j+k];++k);
}
inline int solve()
{
	int ans=0;
	for(register int i=1;i<=n;++i)ans+=n-SA[i]+1-height[i];
	return ans;
}
int main()
{
	read(T);
	while(T--)
	{
		scanf("%s",s+1);
		GetSA();
		write(solve(),'\n');
	}
	return 0;
}